Problem: $\dfrac{dy}{dt}=3t$ and $y(2)=3$. What is $t$ when $y=6$ ? Choose all answers that apply: Choose all answers that apply: (Choice A) A $t=6$ (Choice B) B $t=\sqrt{6}$ (Choice C) C $t=-2$ (Choice D) D $t=-\sqrt{6}$ (Choice E) E $t=2$
Explanation: The differential equation is separable. What does it look like after we separate the variables? $dy=3t\,dt$ Let's integrate both sides of the equation. $\int dy=\int3t\,dt$ What do we get? $y=\dfrac32t^2+C$ What value of $C$ satisfies the initial condition $y(2)=3$ ? Let's substitute $t=2$ and $y=3$ into the equation and solve for $C$. $\begin{aligned} 3&=\dfrac32\cdot2^2+C\\ \\ \\ 3&=6+C\\ \\ C&=-3 \end{aligned}$ Now use this value of $C$ to find $t$ when $y=6$. $\begin{aligned} 6&=\dfrac32t^2-3\\ \\ \\ \\ 9&=\dfrac32t^2\\ \\ \\ \\ 6&=t^2\\ \\ t&=\pm \sqrt6 \end{aligned}$